Partea intreaga
Posted: Tue Feb 17, 2009 8:14 pm
Sa se afle \( a,b,c\in \mathbb{N}^* \) astfel incat \( [\frac{a^2}{2b}]+[\frac{b^2}{2c}]+[\frac{c^2}{2a}]=0 \) .
\( a=b=c=1 \)
Intradevar,
Din \( 0\le \frac{a^2}{2b}<1 \) rezulta \( a^2+1\le 2b \) si analoagele, de unde prin adunare,
\( (a-1)^2+(b-1)^2+(c-1)^2\le 0 \) si de aici concluzia.